Question

1) Zinc reacts with hydrochloric acid according to the reaction equation.How many milliliters of 4.00 M HCl(aq) are required to react with 5.85 g of an ore containing 44.0% Zn(s) by mass?

2) The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation.

2SO2O3(2-)(aq) +I3(-)(aq) --->>> S4O6(2-)(aq) + 3I(-)(aq)

Given that it requires 25.6 mL of 0.460 M Na2S2O3(aq) to titrate a 10.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.

Answer 1

1) Zinc reacts with hydrochloric acid according to the reaction equation.How many milliliters of 4.00 M HCl(aq) are required to react with 5.85 g of an ore containing 44.0% Zn(s) by mass?

Solution :-

Balanced reaction equation

Zn + 2HCl ----- > ZnCl2 + H2

Lets calculate the mass of the Zn using the percent

5.85 g * 44.0 % Zn / 100 % = 2.574 g Zn

Now lets calculate the moles of the Zn

Moles of Zn = mass of Zn / molar mass of Zn

                    = 2.574 g Zn / 65.39 g per mol

                    = 0.03936 mol Zn

Now lets calculate the moles of the HCl using the mole ratio of the Zn and HCl

0.03936 mol Zn * 2 mol HCl / 1 mol Zn = 0.07872 mol HCl

Now lets calculate the volume of the HCl

Volume in liter = moles / molarity

                         = 0.07872 mol / 4.00 mol per L

                         = 0.0197 L

0.0197 L * 1000 ml / 1 L= 19.7 ml

So the volume of HCl needed = 19.7 ml

2) The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation.

2SO2O3(2-)(aq) +I3(-)(aq) --->>> S4O6(2-)(aq) + 3I(-)(aq)

Given that it requires 25.6 mL of 0.460 M Na2S2O3(aq) to titrate a 10.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.

Solution :-

Lets first calculaete the moles of the Na2S2O3

Moles = molarity * volume in liter

           = 0.460 mol per L * 0.0256 L

          = 0.01178 mol Na2S2O3

Now lets calculate the moles of the I3^-

0.01178 mol S2O3^2- * 1 mol I3^- / 2 mol S2O3^2- = 0.00589 mol I3^-

Now lets calculate the molarity of the I3^-

Molarity = moles / volume in liter

10 ml * 1 L / 1000 ml = 0.010 L

So molarity = 0.00589 mol I3^- / 0.010 L

                     = 0.589 M

So the molarity of the I3^- = 0.589 M